Chapter 9 Hypothesis Tests
Solutions: 6. a. H0: μ ≤ 1 Ha: μ > 1 b. Claiming μ > 1 when it is not. This is the error of rejecting the product’s claim when the claim is true. Concluding μ ≤ 1 when it is not. In this case, we miss the fact that the product is not meeting its label specification. H0: μ ≤ 8000 Ha: μ > 8000 b. Research hypothesis to see if the plan increases average sales. The label claim or assumption.

c.

7.

a.

Claiming μ > 8000 when the plan does not increase sales. A mistake could be implementing the plan when it does not help. Concluding μ ≤ 8000 when the plan really would increase sales. This could lead to not implementing a plan that would increase sales. z= x − μ0 = 26.4 − 25 6 / 40 = 1.48

c.

10. a. b.

σ/ n

Upper tail p-value is the area to the right of the test statistic Using normal table with z = 1.48: p-value = 1.0000 - .9306 = .0694 Using Excel: p-value = 1 - NORMSDIST(1.48) = .0694

c. d.

p-value > .01, do not reject H0 Reject H0 if z ≥ 2.33 1.48 < 2.33, do not reject H0

11. a. b.

z=

x − μ0

σ/ n

=

14.15 − 15 3 / 50

= −2.00

Because z < 0, p-value is two times the lower tail area Using normal table with z = -2.00: p-value = 2(.0228) = .0456

9-1

Chapter 9

Using Excel: p-value = 2*NORMSDIST(-2.00) = .0456 c. d. p-value ≤ .05, reject H0 Reject H0 if z ≤ -1.96 or z ≥ 1.96 -2.00 ≤ -1.96, reject H0 15. a. H0: μ ≥ 1056 Ha: μ < 1056 b. z= x − μ0 = 910 − 1056 1600 / 400 = −1.83

σ/ n

Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.83: p-value =.0336 Using Excel: p-value = NORMSDIST(-1.83) = .0336 c. d. p-value ≤ .05, reject H0. Conclude the mean refund of “last minute” filers is less than $1056. Reject H0 if z ≤ -1.645 -1.83 ≤ -1.645, reject H0 23. a.
b. t= x − μ0 s/ n = 14 − 12 4.32 / 25 = 2.31

Degrees of freedom = n – 1 = 24 Upper